There's a famous "World's Hardest Easy Geometry Problem" in http://thinkzone.wlonk.com/MathFun/Triangle.htm, It's solved now. Here is the step by step for the solution.

**
Figure 1
**

From Figure 1, we have

[1] AC = BC

[2] BD = CD

**
Figure 2
**

In Figure 2, parallel to AB, we draw DF

Because [1] and AB||DF, we get

[3] DC = FC

[4] AD = BF

**
Figure 3
**

In Figure 3, we draw line AF.

From [4], and AB=BA, and angle DAB = FBA (=80),

the triangle ABD equals to triangle ABF. So

[5] AF = BD

From [2], AF = CD

From [3], we get

[6] AF = CF

So CAF = 20, EAF = 10, FAB = 60,

[7] DGF = 60

[8] AB = BG = GA

From [5], [7] and [8], we get

[9] DF = FG = GD

**
Figure 4
**

In Figure 4, we draw FH vertical to AC.

From [6] and angle CAF = ACF, we get triangle AFH equals to triangle CFH. So

[10] AH = CH

**
Figure 5
**

In Figure 5, we draw CI.

From [10] and FH vertical to AC, we get

[11] angle DCI = DAI = 10

[12] CI = AI

From [11], we get

[13] angle FCI = 10 = DCI

[13'] angle FCI = FAI

**
Figure 6
**

In Figure 6, extend CI to AB.

From [13], and angle ABC = BAC = 80, we get

line CI vertical to AB,

[14] line CI is the same as line CG

From [12], [13], [14], we get triangle CEI equals to triangle AGI. So

[15] EI = GI

From [12], [15], we get

[16] CG = AE

From [13'], [16], we get triangle CGF equals to triangle AEF. So

[17] EF = FG

**Finally**

From [9], [17], we get

[18] EF = DF

Because DF || AB, so

[19] angle CFD = CBA = 80

From [18], [19], we get

[20] angle DEF = EDF = 50

So angle X = angle DEF - angle AEF

= 50 - (180 - 70 - 60 - 20)

= 50 - 30

= 20